Problem: $f(x, y, z) = (\cos(z), 9xy, x^3)$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-\sin(z), 9x + 9y, 3x^2)$ (Choice B) B $(-\sin(z), 0, 0)$ (Choice C) C $(0, 9x, 0)$ (Choice D) D $(0, 9y, 3x^2)$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y, z) = (f_0(x, y, z), f_1(x, y, z), f_2(x, y, z)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x}, \dfrac{\partial f_2}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y}, \dfrac{\partial f_2}{\partial y} \right) \\ \\ &f_z = \left( \dfrac{\partial f_0}{\partial z}, \dfrac{\partial f_1}{\partial z}, \dfrac{\partial f_2}{\partial z} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $x$, we'll treat $y$ and $z$ as if they were constants. Therefore, $f_x = (0, 9y, 3x^2)$.